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lidiya [134]
3 years ago
5

Consider an ionic compound, MX, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MXis ΔH

∘f=−427kJ/mol. The enthalpy of sublimation of Mis ΔHsub=135kJ/mol. The ionization energy of Mis IE=433kJ/mol. The electron affinity of Xis ΔHEA=−307kJ/mol. (Refer to the hint). The bond energy of X2is BE=175kJ/mol. Determine the lattice energy of MX.
Chemistry
1 answer:
IrinaK [193]3 years ago
5 0

Answer:

The lattice energy of MX is -523.5 kJ/mol.

Explanation:

M(s)+\frac{1}{2}X_2(g)\rightarrow MX(s),\Delta H_f=-427kJ/mol....[1]

M(s)\rightarrow M(g),\Delta H_{sub}=135kJ/mol....[2]

M(g)\rightarrow M^+(g)+1e^-,I.E=433kJ/mol....[3]

X(g)+e^-\rightarrow X^-(g),E.A=-307kJ/mol....[4]

X_2(g)\rightarrow 2X(g), \Delta H_{X-X}=175kJ/mol....[5]

M^+(g)+X^-(g)\rightarrow MX(s),L.E=?....[6]

[1] = [2] + (1/2)[5] + [6] - [3] - [4]

-427kJ/mol = 135kJ/mol +\frac{1}{2}\times 175kJ/mol+L.E- (433kJ/mol) - (-307kJ/mol)

-427kJ/mol=96.5 kJ/mol+L.E

L.E=-523.5 kJ/mol

The lattice energy of MX is -523.5 kJ/mol.

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How can one kg of iron melt more ice than 1 kg lead at 100 °C
Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{iron} = Heat obtained from the iron by the ice

ΔQ_{iron} = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{lead} = Heat obtained from the iron by the ice

ΔQ_{lead} = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

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3 years ago
The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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0.254g lead(ii)ethanoate, on adding excess K2CrO4 solution, gave 0.130g of lead(ii)chromate precipitate. what is the percentage
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Answer:

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Explanation:

All of the Pb⁺² species precipitated as lead(II) cromate, PbCrO₄ (we know this as excess K₂CrO₄ was used).

First we convert 0.130 g of PbCrO₄ into moles, using its molar mass:

  • 0.130 g ÷ 323 g/mol = 4.02x10⁻⁴ mol PbCrO₄

There's 1 Pb⁺² mol per PbCrO₄ mol, so in total 4.02x10⁻⁴ moles of Pb⁺² were in the ethanoate sample.

We <u>convert those 4.02x10⁻⁴ moles of Pb into grams</u>:

  • 4.02x10⁻⁴ mol * 207 g/mol = 0.083 g Pb

Finally we calculate the percentage composition of Pb:

  • 0.083 g Pb / 0.254 g salt * 100% = 32.8%
3 0
3 years ago
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